TITLE: Conjugacy for $p$-adic matrices of finite order
QUESTION [13 upvotes]: Say $p$ is an odd prime, and take two matrices $A,B\in GL_n({\mathbb Z}_p)$ of finite order $m$. Is it true that they are conjugate in $GL_n({\mathbb Z}_p)$ if and only if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$?
Edit: Thank you all for the answers and a very insightful discussion! As the answer is NO, it is important for me to know whether conjugacy in $GL_n({\mathbb F}_p)$ at least implies conjugacy in $GL_n({\mathbb Q}_p)$. I cannot see this from the answers, so I think I will ask this as a separate question.

REPLY [2 votes]: Here is a trial proof for the question over $Q_p$.
Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$).
Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers).
Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$ for $(u,p)=1$. Further note, if $\bar\Phi_u=\prod \bar g$ then $\bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$,
and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split,
in other words this is the minimal polynomial. This follows since the reduction (mod $p$) of the companion matrix of $f$ is itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$.
So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$.
From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. The general Jordan form classifies the conjugacy type over a field, as is $Q_p$.
Note that, $\Phi_3\Phi_6$ and $\Phi_6^2$ give 4x4 matrices with order 6, failing for $p=2$.